I have another question for the problem below, so What is the speed of the car just before it lands safely on the other side? would it be the same 38.6 meters/sec
A 10,000N car comes to a bridge during a storm and finds the bridge washed out. The 650-N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 16.6 m above the river, while the opposite side is a mere 7.90 m above the river. The river itself is a raging torrent 51.4 m wide. How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
This question has already been posted on here but I need to know how they found the initial velocity.
Physics - Damon, Friday, February 8, 2008 at 5:13pm
Horizontal problem:
Speed = constant = U, the initial speed
distance = U t = 51.4 m
Vertical problem:
falls (16.6 - 7.9) = 8.7 meters DOWN
initial speed down = 0
so
8.7 = (1/2) (9.8) t^2
weight has no effect on this problem, only the acceleration of gravity
8.7 = 4.9 t^2
t^2 = 1.77
t = 1.33 seconds to9 fall
same t to reach the other bank so
U = 51.4 meters / 1.33 seconds
= 38.6 meters/sec
Physics - Charlie, Friday, February 8, 2008 at 5:38pm
Thank you so much.
Physics - Damon, Friday, February 8, 2008 at 6:42pm
You are welcome !
To find the speed of the car just before it lands safely on the other side, you need to calculate the initial velocity at which the car should be traveling just as it leaves the cliff. Let's break down the solution step-by-step:
1. To clear the river and land safely on the opposite side, the car needs to travel a horizontal distance of 51.4 meters.
2. Assuming the horizontal speed is constant, we can denote it as "U" (the initial speed).
3. The time taken to cross the river horizontally is the same as the time taken for the car to fall vertically from a height of 8.7 meters. This is because the vertical motion and horizontal motion occur simultaneously.
4. The vertical motion equation for falling objects is: distance = (1/2)gt^2, where "g" is the acceleration due to gravity (9.8 m/s^2) and "t" is the time taken.
5. In this case, the car falls 8.7 meters downwards, so the equation becomes: 8.7 = (1/2)(9.8)t^2.
6. Solving this equation, we find that t^2 = 1.77.
7. Taking the square root of both sides, we get t ≈ 1.33 seconds.
8. Since the time taken to fall vertically is the same as the time taken to reach the other bank horizontally, the time for horizontal motion is also approximately 1.33 seconds.
9. Now, we can calculate the initial speed (U) by dividing the horizontal distance (51.4 meters) by the time taken (1.33 seconds), giving us U ≈ 38.6 m/s.
Therefore, the speed of the car just before it lands safely on the other side is approximately 38.6 meters/second.